Question

我可以选择并显示数据库中的所有图像。现在我想通过id过滤它们。如何通过id从db获取图像？

有没有办法比较AsyncTask中的字符串“imovel_id”和从db中选择的“imovel_id”json响应？

或者通过Android应用程序发送的请求获取字符串“imovel_id”然后通过“imovel_id”选择所有图像会更好吗？

我尝试在下面的代码中从AsyncTask发送字符串，在php请求上获取它，然后使用WHERE clausule的查询，但我没有成功。我得到一个空值。

include_once 'db_connect.php';

$i_id = $_REQUEST["imovel_id"];

$sql = "SELECT * FROM iMoveis WHERE imovel_id='$i_id'";
$result = mysqli_query($mysqli, $sql);

$response = array();
$images = array();


while($row = mysqli_fetch_assoc($result)){

 $images[] = array('images' => $row['img1']);
 $images[] = array('images' => $row['img2']);
 $images[] = array('images' => $row['img3']);
 $images[] = array('images' => $row['img4']);
 $images[] = array('images' => $row['img5']);

}

$response['posts'] = $images;

echo json_encode($response, JSON_UNESCAPED_SLASHES);

php脚本

{{1}}

在json数组中添加id然后在parseResult方法中将其与textview中的字符串“imovel_id”进行比较可能是一个解决方案吗？

我认为通过nameValuePairs发送字符串会很容易，因为我已经完成并在查询中使用WHERE进行过滤，但似乎这样做不那么容易。

提前致谢。

Answer 1

我在AsyncTask中做了一些更改

public class AsyncHttpTask extends AsyncTask<String, Void, Integer> {

    @Override
    protected Integer doInBackground(String... params) {

        ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
        nameValuePairs.add(new BasicNameValuePair("imovel_id", i_id));


        Integer result = 0;
        try {
            // Create Apache HttpClient
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://meuwebsite.com/panel/json_images.php");
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse httpResponse = httpclient.execute(httppost);

            int statusCode = httpResponse.getStatusLine().getStatusCode();

            // 200 represents HTTP OK
            if (statusCode == 200) {
                String response = streamToString(httpResponse.getEntity().getContent());
                parseResult(response);
                result = 1; // Successful
            } else {
                result = 0; //"Failed
            }
        } catch (Exception e) {
            Log.d(TAG, e.getLocalizedMessage());
        }

        return result;
    }

    @Override
    protected void onPostExecute(Integer result) {
        // Download complete. Lets update UI

        if (result == 1) {
            mGridAdapter.setGridData(mGridData);
        } else {
            Toast.makeText(GridViewActivity.this, "Failed to fetch data!", Toast.LENGTH_SHORT).show();
        }

        //Hide progressbar
        mProgressBar.setVisibility(View.GONE);
    }
}


String streamToString(InputStream stream) throws IOException {
    BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(stream));
    String line;
    String result = "";
    while ((line = bufferedReader.readLine()) != null) {
        result += line;
    }

    // Close stream
    if (null != stream) {
        stream.close();
    }
    return result;
}

/**
 * Parsing the feed results and get the list
 *
 * @param result
 */
private void parseResult(String result) {
    try {
        JSONObject response = new JSONObject(result);
        JSONArray posts = response.optJSONArray("posts");
        GridItem item;
        for (int i = 0; i < posts.length(); i++) {
            JSONObject post = posts.optJSONObject(i);
            item = new GridItem();
            item.setImage(post.getString("images"));
            mGridData.add(item);
        }
    } catch (JSONException e) {
        e.printStackTrace();
    }
}

还将$i_id = $_REQUEST["imovel_id"];更改为$i_id = $_POST["imovel_id"];

Android + json通过id获取图像

1 个答案: